Problem: $x+2xy-y^2=2$ Find the value of $\dfrac{dy}{dx}$ at the point $(2,4)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{2}$ (Choice B) B $\dfrac{1}{2}$ (Choice C) C $-\dfrac{9}{4}$ (Choice D) D $\dfrac{9}{4}$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $x+2xy-y^2=2$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} x+2xy-y^2&=2 \\\\ \dfrac{d}{dx}(x+2xy-y^2)&=\dfrac{d}{dx}(2) \\\\ \dfrac{d}{dx}(x)+2\dfrac{d}{dx}(xy)-\dfrac{d}{dx}(y^2)&=0 \\\\ 1+2\left(1\cdot y+x\cdot\dfrac{dy}{dx}\right)-2y\cdot\dfrac{dy}{dx}&=0 \\\\ 1+2y+2x\cdot\dfrac{dy}{dx}-2y\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 1+2y+2x\cdot\dfrac{dy}{dx}-2y\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(2x-2y)&=-(1+2y) \\\\ \dfrac{dy}{dx}&=-\dfrac{1+2y}{2x-2y} \end{aligned}$ Now we can plug the point $(2,4)$ into the expression for $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=-\dfrac{1+2y}{2x-2y} \\\\ &=-\dfrac{1+2(4)}{2(2)-2(4)} \gray{x=2,\,\,y=4} \\\\ &=-\dfrac{9}{-4} \\\\ &=\dfrac{9}{4} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at the point $(2,4)$ is $\dfrac{9}{4}$.